, 2019

The internal bisectrix $AI$ and the perpendicular bisectrix $MN$ of the side $BC$ cross at the midpoint $K$ of the arc $BEC$. It is a fact that the circle $BIC$ is centred at $K$.

Let now the line $MP$ cross the circle $BIC$ again at $Q$, to infer that the arcs $BL$ and $CQ$ of this circle have equal angular spans, so $L$ and $Q$ are reflexions of one another in the perpendicular bisectrix $KMN$ of the chord $BC$.
Project $Q$ orthogonally to $Q'$ on $BC$ and refer to standard notation in the triangle $ABC$: $a$, $b$, $c$ denote the lengths of the sides $BC$, $CA$, $AB$, respectively, $s = (a + b + c)/2$ denotes its semiperimeter, $r$ its inradius, and $S$ its area. With reference to standard formulae, write $CQ' = BD = s - b$ and
$$QQ' = DL = \frac{DB \cdot DC}{DI} = \frac{(s-b)(s-c)}{r} = \frac{s}{s-a},$$
to infer that $Q$ is the $A$-excentre of the triangle $ABC$, so it lies on the line $AIK$.
Finally, write $\angle PQA = \angle PQI = \angle PLI = \angle ELI = \angle EAI = \angle DAQ$, to conclude that the lines $AD$ and $MP$ are indeed parallel.